In the given circuit,what is the voltage betw | vedclass

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ in the path from $Q$ to $P$:Starting from point $Q$,moving towards $P$ against the direction of current $

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$5$

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ in the path from $Q$ to $P$:Starting from point $Q$,moving towards $P$ against the direction of current $I = 2 \, A$ through the resistor $R$ (which is not explicitly given,but the potential drop is determined by the loop):Actually,looking at the circuit,the current $2 \, A$ flows from $Q$ to $P$. The potential at $P$ is $15 \, V$ relative to ground.The potential at $Q$ can be found by considering the path from $P$ to $Q$:$V_P - I \cdot R_{total} = V_Q$ is not applicable as $R$ is unknown. However,we can use the loop equation:$V_P - V_Q = I \cdot R_{PQ}$.Given the diagram,the potential at $P$ is $15 \, V$.The current $2 \, A$ flows from $Q$ to $P$.Using the path from $P$ to $Q$: $V_P - I \cdot R = V_Q$.Wait,the provided solution uses the equation: $- 5 	imes 2 - V_{PQ} + 15 = 0$.This implies $V_P - V_Q = 15 - (2 	imes 5) = 15 - 10 = 5 \, V$.Thus,the potential difference $V_{PQ} = 5 \, V$.

In the given circuit,what is the voltage between $P$ and $Q$ in $V$?

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